Question

A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of \(2.50 \mathrm{kg} \cdot \mathrm{m}^{2}\) when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to \(1.60 \mathrm{kg} \cdot \mathrm{m}^{2} ?\)

Short answer

Answer: The final angular velocity of the skater is 15.625 rad/s.

Step-by-step solution

Understand the conservation of angular momentum

Angular momentum is a conserved quantity in a system when there is no net external torque acting on it. In other words, angular momentum is constant. The equation for angular momentum (L) is given by: L = I * ω where I is the rotational inertia and ω is the angular velocity.

Write the initial and final angular momentum equations

The initial angular momentum (L1) is given by the initial rotational inertia (I1) and initial angular velocity (ω1). The final angular momentum (L2) is given by the final rotational inertia (I2) and final angular velocity (ω2). We have: L1 = I1 * ω1 L2 = I2 * ω2

Apply the conservation of angular momentum

Since angular momentum is conserved, we have: L1 = L2 I1 * ω1 = I2 * ω2

Solve for the final angular velocity (ω2)

We know the initial angular velocity (ω1), the initial rotational inertia (I1), and the final rotational inertia (I2). We need to find the final angular velocity (ω2). We can solve for ω2 in the equation: I1 * ω1 = I2 * ω2 Rearranging the equation to solve for ω2, we get: ω2 = (I1 * ω1) / I2

Substitute the given values and solve for ω2

Now we can substitute the given values: I1 = 2.50 kg·m² I2 = 1.60 kg·m² ω1 = 10.0 rad/s ω2 = (2.50 kg·m² * 10.0 rad/s) / 1.60 kg·m² Calculating, we find ω2: ω2 = 15.625 rad/s The skater's angular velocity after she pulls her arms in is 15.625 rad/s.